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Let $G$ be a group, and let $N$ be a normal subgroup of $G$. I know that the natural projection homomorphism is a surjective homomorphism from $G$ onto $G/N$. If I choose a particular representative, say $xN$ and map every element in the coset $xN$ of $N$ in $G$ to $x$, will I get a "well-defined" homomorphism from $G$ to $G$ with kernel $N$. (Obviously this map is not onto, unless $N$ is trivial.)

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    $\begingroup$ When $G\approx N\times H$, then yes, there is a map. Basically, whenever there is, in any coset, a distinguished representative, you may be in luck. $\endgroup$ – Arthur Jul 13 '15 at 19:57
  • $\begingroup$ The theorem of Schur-Zassenhaus is a classic result giving a sufficient condition for such a homomorphism to exist: when the order $|N|$ of the normal subgroup $N$ is coprime to its index $|G/N|$. $\endgroup$ – j.p. Jul 14 '15 at 9:46
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$N$ need not be the identity subgroup. There is another important situation where this occurs:

Suppose $G = HN$, where $N \lhd G$, and $H \cap N = \{e\}$.

Then $G/N = HN/N \cong H/(H \cap N) \cong H$.

In this situation, for any $g \in G$, we have a unique representation $g = hn$, for some $h \in H$, and $n \in N$, so we can define:

$\phi: G/N \to G$ by $\phi(gN) = \phi(hnN) = \phi(hN) = h$ (verify this is well-defined!).

Note that if $\pi: G \to G/N$ is the map $g \mapsto gN$, we have:

$\pi \circ \phi = \text{id}_{G/N}$. With such a happy state of affairs, $G$ is said to be right-split (over $N$), or: the semi-direct product of $N$ by $H$. In the special case that $H \lhd G$ as well, we obtain the direct product, as mentioned in an earlier comment.

Note that if we take $N = \{e\}$, which is always normal in $G$, we get the situation described by Eoin, which admittedly is not very interesting.

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  • $\begingroup$ In this construction one can replace the complement $H$ of $N$ by any of its conjugates (but not all possible complements have to be conjugated). $\endgroup$ – j.p. Jul 14 '15 at 9:30
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The answer is no; the reason is that, in general, you can't define a group homomorphism from $G/N$ to $G$ sending $xN$ to $x$. Consider the following example: $G=\{\pm1,\pm{i},\pm{j},\pm{k}\}$, the quaternion group, and $N=\{\pm1,\pm{i}\}$. Thus $G/N=\{1N,jN\}$, and you send $jN$ (of order 2) to $j$ (of order 4), a contradiction.

On the other hand, you may send $jN$ in $G/N$, to $-1$, in $G$, the only element of $G$ of order 2; in this case you have a group endomorphism $f$ of $G$ sending $j$ to $-1$, and $i$ to 1. Hence $\mathrm{Ker}(f)=\{\pm1,\pm{i}\}=N$; so a related interesting question could be: Is every normal subgroup $N$ of a group $G$ the kernel of an endomorphism?

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  • $\begingroup$ Additionally: If you take $N=\{\pm 1\}$ then $G$ does not even have any subgroup isomorphic to $G/N$. $\endgroup$ – j.p. Jul 14 '15 at 9:25
  • $\begingroup$ Perfect, you are right! Which is interesting is that $G$ is not a non-trivial semidirect product! $\endgroup$ – arienda Jul 14 '15 at 10:17
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No if $N$ anything other than just the identity. In that case you would have two distinct elements of $G$ such that $g_1 N=g_2 N$ and you would thus have $f(g_1 N)=f(g_2 N)=g_1=g_2$, a contradiction.

However, and I'm not sure if this bares any relevance to you, but in the proof of the first isomorphism theorem, we see that if $N$ is the kernel of a surjective homomorphism $\phi: G\to \bar G$, then there is an isomorphism $\pi:G/N\to \bar G$ defined by $\pi (gN)=\phi(g)$ for all $gN\in G/N$.

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